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Use the power series for f(x) = 1/(1-x) to write the function g(x) = x^2/(2 - x^2) as a power series. Write your answer as a summation with a lower index of n = 0, and simplify the summation to a single variable.


Would really appreciate the help

Verified Answer :

LammettHash

\(\dfrac{x^2}{2-x^2}=\dfrac{x^2}2\cdot\dfrac1{1-\frac{x^2}2}\)

Since

\(\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n\)

for \(|x|<1|\), the power series for the rightmost rational expression is

\(\dfrac1{1-\frac{x^2}2}=\displaystyle\sum_{n=0}^\infty\left(\frac{x^2}2\right)^n=\sum_{n=0}^\infty \frac{x^{2n}}{2^n}\)

which is applicable for \(\left|\dfrac{x^2}2\right|<1\), or \(|x|<\sqrt2\).

Distributing the other rational expression gives

\(\dfrac{x^2}{2-x^2}=\displaystyle\frac{x^2}2\sum_{n=0}^\infty\frac{x^{2n}}{2^n}=\sum_{n=0}^\infty\frac{x^{2n+2}}{2^{n+1}}\)

I'm not sure what is meant by "simplify the summation to a single variable", but you can rewrite this as

\(\displaystyle\sum_{n=0}^\infty\frac{x^{2(n+1)}}{2^{n+1}}=\sum_{n=0}^\infty\left(\frac{x^2}2\right)^{n+1}\)

which could be interpreted as a simpler form, but that's really in the eye of the beholder...

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